Database System Concepts, 5
th
Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
Chapter 2: Relational Model
Chapter 2: Relational Model
©Silberschatz, Korth and Sudarshan2.2Database System Concepts - 5th Edition, Oct 5, 2006
Chapter 2: Relational Model
Chapter 2: Relational Model
Structure of Relational Databases
Fundamental Relational-Algebra-Operations
Additional Relational-Algebra-Operations
Extended Relational-Algebra-Operations
Null Values
Modification of the Database
©Silberschatz, Korth and Sudarshan2.3Database System Concepts - 5th Edition, Oct 5, 2006
Example of a Relation
Example of a Relation
©Silberschatz, Korth and Sudarshan2.4Database System Concepts - 5th Edition, Oct 5, 2006
Basic Structure
Basic Structure
Formally, given sets D
1
, D
2
, …. D
n
a relation ris a subset of
D
1
x D
2
x … x D
n
Thus, a relation is a set of n-tuples (a
1
,a
2
, …, a
n
) where each a
i
D
i
Example: If
customer_name = {Jones, Smith, Curry, Lindsay, …}
/* Set of all customer names */
customer_street = {Main, North, Park, …} /* set of all street names*/
customer_city = {Harrison, Rye, Pittsfield, …} /* set of all city names */
Then r= { (Jones, Main, Harrison),
(Smith, North, Rye),
(Curry, North, Rye),
(Lindsay, Park, Pittsfield) }
is a relation over
customer_name x customer_street x customer_city
©Silberschatz, Korth and Sudarshan2.5Database System Concepts - 5th Edition, Oct 5, 2006
Attribute Types
Attribute Types
Each attribute of a relation has a name
The set of allowed values for each attribute is called the domain of the
attribute
Attribute values are (normally) required to be atomic; that is, indivisible
E.g. the value of an attribute can be an account number,
but cannot be a set of account numbers
Domain is said to be atomic if all its members are atomic
The special value null is a member of every domain
The null value causes complications in the definition of many operations
We shall ignore the effect of null values in our main presentation
and consider their effect later
©Silberschatz, Korth and Sudarshan2.6Database System Concepts - 5th Edition, Oct 5, 2006
Relation Schema
Relation Schema
A
1
, A
2
, …, A
n
are attributes
R= (A
1
, A
2
, …, A
n
) is a relation schema
Example:
Customer_schema = (customer_name, customer_street, customer_city)
r(R) denotes a relation r on the relation schema R
Example:
customer (Customer_schema)
©Silberschatz, Korth and Sudarshan2.7Database System Concepts - 5th Edition, Oct 5, 2006
Relation Instance
Relation Instance
The current values (relation instance) of a relation are specified by
a table
An element tof ris a tuple, represented by a row in a table
Jones
Smith
Curry
Lindsay
customer_name
Main
North
North
Park
customer_street
Harrison
Rye
Rye
Pittsfield
customer_city
customer
attributes
(or columns)
tuples
(or rows)
©Silberschatz, Korth and Sudarshan2.8Database System Concepts - 5th Edition, Oct 5, 2006
Relations are Unordered
Relations are Unordered
Order of tuples is irrelevant (tuples may be stored in an arbitrary order)
Example: account relation with unordered tuples
©Silberschatz, Korth and Sudarshan2.9Database System Concepts - 5th Edition, Oct 5, 2006
Database
Database
A database consists of multiple relations
Information about an enterprise is broken up into parts, with each relation
storing one part of the information
account : stores information about accounts
depositor : stores information about which customer
owns which account
customer : stores information about customers
Storing all information as a single relation such as
bank(account_number, balance, customer_name, ..)
results in
repetition of information
e.g.,if two customers own an account (What gets repeated?)
the need for null values
e.g., to represent a customer without an account
Normalization theory (Chapter 7) deals with how to design relational schemas
©Silberschatz, Korth and Sudarshan2.10Database System Concepts - 5th Edition, Oct 5, 2006
The
The customer
customer Relation
Relation
©Silberschatz, Korth and Sudarshan2.11Database System Concepts - 5th Edition, Oct 5, 2006
The
The depositor
depositor Relation
Relation
©Silberschatz, Korth and Sudarshan2.12Database System Concepts - 5th Edition, Oct 5, 2006
Keys
Keys
Let K R
K is a superkey of Rif values for Kare sufficient to identify a unique tuple of
each possible relation r(R)
by “possible r we mean a relation rthat could exist in the enterprise we
are modeling.
Example: {customer_name, customer_street} and
{customer_name}
are both superkeys of Customer, if no two customers can possibly have
the same name
In real life, an attribute such as customer_id would be used instead of
customer_name to uniquely identify customers, but we omit it to keep
our examples small, and instead assume customer names are unique.
©Silberschatz, Korth and Sudarshan2.13Database System Concepts - 5th Edition, Oct 5, 2006
Keys (Cont.)
Keys (Cont.)
Kis a candidate key if Kis minimal
Example: {customer_name} is a candidate key for Customer, since it
is a superkey and no subset of it is a superkey.
Primary key: a candidate key chosen as the principal means of
identifying tuples within a relation
Should choose an attribute whose value never, or very rarely,
changes.
E.g. email address is unique, but may change
©Silberschatz, Korth and Sudarshan2.14Database System Concepts - 5th Edition, Oct 5, 2006
Foreign Keys
Foreign Keys
A relation schema may have an attribute that corresponds to the primary
key of another relation. The attribute is called a foreign key.
E.g. customer_name and account_number attributes of depositor are
foreign keys to customer and account respectively.
Only values occurring in the primary key attribute of the referenced
relation may occur in the foreign key attribute of the referencing
relation.
Schema diagram
©Silberschatz, Korth and Sudarshan2.15Database System Concepts - 5th Edition, Oct 5, 2006
Query Languages
Query Languages
Language in which user requests information from the database.
Categories of languages
Procedural
Non-procedural, or declarative
“Pure” languages:
Relational algebra
Tuple relational calculus
Domain relational calculus
Pure languages form underlying basis of query languages that people
use.
©Silberschatz, Korth and Sudarshan2.16Database System Concepts - 5th Edition, Oct 5, 2006
Relational Algebra
Relational Algebra
Procedural language
Six basic operators
select:
σ
project:
union:
set difference:
Cartesian product: x
rename:
ρ
The operators take one or two relations as inputs and produce a new
relation as a result.
©Silberschatz, Korth and Sudarshan2.17Database System Concepts - 5th Edition, Oct 5, 2006
Select Operation
Select Operation
Example
Example
Relation r A B C D
α
α
β
β
α
β
β
β
1
5
12
23
7
7
3
10
σ
A=B ^ D > 5
(r)
A B C D
α
β
α
β
1
23
7
10
©Silberschatz, Korth and Sudarshan2.18Database System Concepts - 5th Edition, Oct 5, 2006
Select Operation
Select Operation
Notation:
σ
p
(r)
pis called the selection predicate
Defined as:
σ
p
(r) = {t| trand p(t)}
Where pis a formula in propositional calculus consisting of terms
connected by : (and), (or), ¬(not)
Each term is one of:
<attribute> op <attribute> or <constant>
where op is one of: =, , >, . <.
Example of selection:
σ
branch_name=“Perryridge”
(account)
©Silberschatz, Korth and Sudarshan2.19Database System Concepts - 5th Edition, Oct 5, 2006
Project Operation
Project Operation
Example
Example
Relation r:A B C
α
α
β
β
10
20
30
40
1
1
1
2
A C
α
α
β
β
1
1
1
2
=
A C
α
β
β
1
1
2
A,C
(r)
©Silberschatz, Korth and Sudarshan2.20Database System Concepts - 5th Edition, Oct 5, 2006
Project Operation
Project Operation
Notation:
where A
1
, A
2
are attribute names and ris a relation name.
The result is defined as the relation of kcolumns obtained by erasing
the columns that are not listed
Duplicate rows removed from result, since relations are sets
Example: To eliminate the branch_name attribute of account
account_number, balance
(account)
)(
,,,
21 r
k
AAA K
©Silberschatz, Korth and Sudarshan2.21Database System Concepts - 5th Edition, Oct 5, 2006
Union Operation
Union Operation
Example
Example
Relations r, s:
r s:
A B
α
α
β
1
2
1
A B
α
β
2
3
r
s
A B
α
α
β
β
1
2
1
3
©Silberschatz, Korth and Sudarshan2.22Database System Concepts - 5th Edition, Oct 5, 2006
Union Operation
Union Operation
Notation: rs
Defined as:
rs= {t| tror ts}
For rsto be valid.
1. r, s must have the same arity (same number of attributes)
2. The attribute domains must be compatible (example: 2
nd
column
of rdeals with the same type of values as does the 2
nd
column of s)
Example: to find all customers with either an account or a loan
customer_name
(depositor)
customer_name
(borrower)
©Silberschatz, Korth and Sudarshan2.23Database System Concepts - 5th Edition, Oct 5, 2006
Set Difference Operation
Set Difference Operation
Example
Example
Relations r, s:
r – s:
A B
α
α
β
1
2
1
A B
α
β
2
3
r
s
A B
α
β
1
1
©Silberschatz, Korth and Sudarshan2.24Database System Concepts - 5th Edition, Oct 5, 2006
Set Difference Operation
Set Difference Operation
Notation r – s
Defined as:
r – s = {t| trand t s}
Set differences must be taken between compatible
relations.
rand smust have the same arity
attribute domains of r and s must be compatible
©Silberschatz, Korth and Sudarshan2.25Database System Concepts - 5th Edition, Oct 5, 2006
Cartesian
Cartesian-
-Product Operation
Product Operation
Example
Example
Relations r, s:
rxs:
A B
α
β
1
2
A B
α
α
α
α
ββββ
1
1
1
1
2
2
2
2
C D
α
ββγα
ββγ
10
10
20
10
10
10
20
10
E
a
a
b
b
a
a
b
b
C D
α
ββγ
10
10
20
10
E
a
a
b
b
r
s
©Silberschatz, Korth and Sudarshan2.26Database System Concepts - 5th Edition, Oct 5, 2006
Cartesian
Cartesian-
-Product Operation
Product Operation
Notation r xs
Defined as:
rx s= {t q |t r and q s}
Assume that attributes of r(R) and s(S) are disjoint. (That is, RS=
).
If attributes of r(R) and s(S) are not disjoint, then renaming must be
used.
©Silberschatz, Korth and Sudarshan2.27Database System Concepts - 5th Edition, Oct 5, 2006
Composition of Operations
Composition of Operations
Can build expressions using multiple operations
Example: σ
A=C
(r x s)
r x s
σ
A=C
(r x s)
A B
α
α
α
α
ββββ
1
1
1
1
2
2
2
2
C D
α
ββ
γα
ββγ
10
10
20
10
10
10
20
10
E
a
a
b
b
a
a
b
b
A B C D E
α
ββ
1
2
2
α
ββ
10
10
20
a
a
b
©Silberschatz, Korth and Sudarshan2.28Database System Concepts - 5th Edition, Oct 5, 2006
Rename Operation
Rename Operation
Allows us to name, and therefore to refer to, the results of relational-
algebra expressions.
Allows us to refer to a relation by more than one name.
Example:
ρ
x
(E)
returns the expression Eunder the name X
If a relational-algebra expression Ehas arity n, then
returns the result of expression Eunder the name X, and with the
attributes renamed to A1 , A
2
, …., A
n
.
)(
),...,,(
21 E
n
AAAx
ρ
©Silberschatz, Korth and Sudarshan2.29Database System Concepts - 5th Edition, Oct 5, 2006
Banking Example
Banking Example
branch (branch_name, branch_city, assets)
customer (customer_name, customer_street, customer_city)
account (account_number, branch_name, balance)
loan (loan_number, branch_name, amount)
depositor (customer_name, account_number)
borrower (customer_name, loan_number)
©Silberschatz, Korth and Sudarshan2.30Database System Concepts - 5th Edition, Oct 5, 2006
Example Queries
Example Queries
Find all loans of over $1200
Find the loan number for each loan of an amount greater than
$1200
σ
amount > 1200
(loan)
loan_number
(σ
amount > 1200
(loan))
Find the names of all customers who have a loan, an account, or both,
from the bank
customer_name
(borrower)
customer_name
(depositor)
©Silberschatz, Korth and Sudarshan2.31Database System Concepts - 5th Edition, Oct 5, 2006
Example Queries
Example Queries
Find the names of all customers who have a loan at the Perryridge
branch.
Find the names of all customers who have a loan at the
Perryridge branch but do not have an account at any branch of
the bank.
customer_name
(σ
branch_name = “Perryridge”
(σ
borrower.loan_number = loan.loan_number
(borrower x loan))) –
customer_name
(depositor)
customer_name
(
σ
branch_name=“Perryridge
(
σ
borrower.loan_number = loan.loan_number
(borrower x
loan)))
©Silberschatz, Korth and Sudarshan2.32Database System Concepts - 5th Edition, Oct 5, 2006
Example Queries
Example Queries
Find the names of all customers who have a loan at the Perryridge branch.
Query 2
customer_name
(σ
loan.loan_number = borrower.loan_number
(
(σ
branch_name = “Perryridge
(loan)) x borrower))
Query 1
customer_name
(σ
branch_name = “Perryridge”
(
σ
borrower.loan_number = loan.loan_number
(borrower x loan)))
©Silberschatz, Korth and Sudarshan2.33Database System Concepts - 5th Edition, Oct 5, 2006
Example Queries
Example Queries
©Silberschatz, Korth and Sudarshan2.34Database System Concepts - 5th Edition, Oct 5, 2006
Example Queries
Example Queries
Find the largest account balance
Strategy:
Find those balances that are not the largest
Rename account relation as d so that we can compare each
account balance with all others
Use set difference to find those account balances that were not found
in the earlier step.
The query is:
balance
(account) -
account.balance
(σ
account.balance < d.balance
(account x
ρ
d
(account)))
©Silberschatz, Korth and Sudarshan2.35Database System Concepts - 5th Edition, Oct 5, 2006
Formal Definition
Formal Definition
A basic expression in the relational algebra consists of either one of the
following:
A relation in the database
A constant relation
Let E
1
and E
2
be relational-algebra expressions; the following are all
relational-algebra expressions:
E
1
E
2
E
1
E
2
E
1
x E
2
σ
p
(E
1
), Pis a predicate on attributes in E
1
s
(E
1
), Sis a list consisting of some of the attributes in E
1
ρ
x
(E
1
), x is the new name for the result of E
1
©Silberschatz, Korth and Sudarshan2.36Database System Concepts - 5th Edition, Oct 5, 2006
Additional Operations
Additional Operations
We define additional operations that do not add any power to the
relational algebra, but that simplify common queries.
Set intersection
Natural join
Division
Assignment
©Silberschatz, Korth and Sudarshan2.37Database System Concepts - 5th Edition, Oct 5, 2006
Set
Set-
-Intersection Operation
Intersection Operation
Notation: rs
Defined as:
rs= { t | trand ts}
Assume:
r, shave the same arity
attributes of rand sare compatible
Note: rs= r (rs)
©Silberschatz, Korth and Sudarshan2.38Database System Concepts - 5th Edition, Oct 5, 2006
Set
Set-
-Intersection Operation
Intersection Operation
Example
Example
Relation r, s:
rs
A B
α
α
β
1
2
1
A B
α
β2
3
r s
A B
α2
©Silberschatz, Korth and Sudarshan2.39Database System Concepts - 5th Edition, Oct 5, 2006
Notation: r s
Natural
Natural-
-Join Operation
Join Operation
Let rand sbe relations on schemas Rand Srespectively.
Then, r s is a relation on schema R Sobtained as follows:
Consider each pair of tuples t
r
from rand t
s
from s.
If t
r
and t
s
have the same value on each of the attributes in RS, add
a tuple tto the result, where
thas the same value as t
r
on r
thas the same value as t
s
on s
Example:
R= (A, B, C, D)
S= (E, B, D)
Result schema = (A, B, C, D, E)
r s is defined as:
r.A, r.B, r.C, r.D, s.E
(
σ
r.B = s.B
r.D = s.D
(r x s))
©Silberschatz, Korth and Sudarshan2.40Database System Concepts - 5th Edition, Oct 5, 2006
Natural Join Operation
Natural Join Operation
Example
Example
Relations r, s:
A B
α
βγαδ
1
2
4
1
2
C D
αγβγβ
a
a
b
a
b
B
1
3
1
2
3
D
a
a
a
b
b
E
α
βγδ
r
A B
α
α
α
αδ
1
1
1
1
2
C D
α
αγγβ
a
a
a
a
b
E
αγαγδ
s
r s
©Silberschatz, Korth and Sudarshan2.41Database System Concepts - 5th Edition, Oct 5, 2006
Division Operation
Division Operation
Notation:
Suited to queries that include the phrase “for all”.
Let rand sbe relations on schemas Rand Srespectively
where
R= (A
1
, …, A
m
, B
1
, …, B
n
)
S= (B
1
, …, B
n
)
The result of r ÷s is a relation on schema
RS = (A
1
, …, A
m
)
r ÷s= { t| t
R-S
(r) u s ( tu r ) }
Where tu means the concatenation of tuples tand uto
produce a single tuple
r ÷s
©Silberschatz, Korth and Sudarshan2.42Database System Concepts - 5th Edition, Oct 5, 2006
Division Operation
Division Operation
Example
Example
Relations r, s:
r÷s:A
B
α
β
1
2
A B
α
α
α
βγδδδ
β
1
2
3
1
1
1
3
4
6
1
2
r
s
©Silberschatz, Korth and Sudarshan2.43Database System Concepts - 5th Edition, Oct 5, 2006
Another Division Example
Another Division Example
A B
α
α
α
ββγγγ
a
a
a
a
a
a
a
a
C D
αγγγγγγβ
a
a
b
a
b
a
b
b
E
1
1
1
1
3
1
1
1
Relations r, s:
r÷s:
D
a
b
E
1
1
A B
αγ
a
a
C
γγ
r
s
©Silberschatz, Korth and Sudarshan2.44Database System Concepts - 5th Edition, Oct 5, 2006
Division Operation (Cont.)
Division Operation (Cont.)
Property
Let q = r ÷s
Then qis the largest relation satisfying qx sr
Definition in terms of the basic algebra operation
Let r(R) and s(S) be relations, and let S R
r÷s= R-S (r ) R-S ( ( R-S (r ) x s ) – R-S,S(r ))
To see why
R-S,S (r) simply reorders attributes of r
R-S (r) x s : all possible pairs between R-S (r ) and s
R-S (r) x s R-S,S(r): those pairs that do not qualify, i.e., do not
appear in r(R).
R-S (R-S (r) x s ) – R-S,S(r) ) gives those tuples t in
R-S (r ) such that for some tuple u s, tu
r.
©Silberschatz, Korth and Sudarshan2.45Database System Concepts - 5th Edition, Oct 5, 2006
Assignment Operation
Assignment Operation
The assignment operation () provides a convenient way to express
complex queries.
Write query as a sequential program consisting of
a series of assignments
followed by an expression whose value is displayed as a result of
the query.
Assignment must always be made to a temporary relation variable.
Example: Write r÷sas
temp1
R-S
(r )
temp2
R-S
((temp1 x s ) –
R-S,S
(r ))
result = temp1 temp2
The result to the right of the is assigned to the relation variable on
the left of the .
May use variable in subsequent expressions.
©Silberschatz, Korth and Sudarshan2.46Database System Concepts - 5th Edition, Oct 5, 2006
Bank Example Queries
Bank Example Queries
Find the names of all customers who have a loan and an account at
bank.
customer_name
(borrower)
customer_name
(depositor)
Find the name of all customers who have a loan at the bank and the
loan amount
customer_name, loan_number, amount
(borrower loan)
©Silberschatz, Korth and Sudarshan2.47Database System Concepts - 5th Edition, Oct 5, 2006
Query 1
customer_name (
σ
branch_name = “Downtown
(depositor account ))
customer_name (
σ
branch_name = “Uptown
(depositor account))
Query 2
customer_name, branch_name (depositor account)
÷
ρ
temp(branch_name
)
({(“Downtown” ), (“Uptown” )})
Note that Query 2 uses a constant relation.
Bank Example Queries
Bank Example Queries
Find all customers who have an account from at least the “Downtown”
and the Uptown” branches.
©Silberschatz, Korth and Sudarshan2.48Database System Concepts - 5th Edition, Oct 5, 2006
Find all customers who have an account at all branches located in
Brooklyn city.
Bank Example Queries
Bank Example Queries
customer_name, branch_name
(depositor account)
÷
branch_name
(σ
branch_city = “Brooklyn”
(branch))
©Silberschatz, Korth and Sudarshan2.49Database System Concepts - 5th Edition, Oct 5, 2006
Extended Relational
Extended Relational-
-Algebra
Algebra-
-Operations
Operations
Generalized Projection
Aggregate Functions
Outer Join
©Silberschatz, Korth and Sudarshan2.50Database System Concepts - 5th Edition, Oct 5, 2006
Generalized Projection
Generalized Projection
Extends the projection operation by allowing arithmetic functions to be
used in the projection list.
Eis any relational-algebra expression
Each of F
1
, F
2
, …, F
n
are are arithmetic expressions involving
constants and attributes in the schema of E.
Given relation credit_info(customer_name, limit, credit_balance), find
how much more each person can spend:
customer_name, limit – credit_balance (credit_info)
)
( ,...,,
21
E
n
FFF
©Silberschatz, Korth and Sudarshan2.51Database System Concepts - 5th Edition, Oct 5, 2006
Aggregate Functions and Operations
Aggregate Functions and Operations
Aggregation function takes a collection of values and returns a single
value as a result.
avg: average value
min: minimum value
max: maximum value
sum: sum of values
count: number of values
Aggregate operation in relational algebra
Eis any relational-algebra expression
G
1
, G
2
…, G
n
is a list of attributes on which to group (can be
empty)
Each F
i
is an aggregate function
Each A
i
is an attribute name
)(
)(,,(),(,,,
221121
E
nnn
AFAFAFGGG KK
ϑ
©Silberschatz, Korth and Sudarshan2.52Database System Concepts - 5th Edition, Oct 5, 2006
Aggregate Operation
Aggregate Operation
Example
Example
Relation r:
A B
α
α
β
β
α
β
β
β
C
7
7
3
10
g
sum(c)
(r) sum(c )
27
©Silberschatz, Korth and Sudarshan2.53Database System Concepts - 5th Edition, Oct 5, 2006
Aggregate Operation
Aggregate Operation
Example
Example
Relation account grouped by branch-name:
branch_name
g
sum(balance)
(
account
)
branch_name account_number balance
Perryridge
Perryridge
Brighton
Brighton
Redwood
A-102
A-201
A-217
A-215
A-222
400
900
750
750
700
branch_name sum(balance)
Perryridge
Brighton
Redwood
1300
1500
700
©Silberschatz, Korth and Sudarshan2.54Database System Concepts - 5th Edition, Oct 5, 2006
Aggregate Functions (Cont.)
Aggregate Functions (Cont.)
Result of aggregation does not have a name
Can use rename operation to give it a name
For convenience, we permit renaming as part of aggregate
operation
branch_name
g
sum(balance) as sum_balance
(
account
)
©Silberschatz, Korth and Sudarshan2.55Database System Concepts - 5th Edition, Oct 5, 2006
Outer Join
Outer Join
An extension of the join operation that avoids loss of information.
Computes the join and then adds tuples form one relation that does
not match tuples in the other relation to the result of the join.
Uses null values:
null
signifies that the value is unknown or does not exist
All comparisons involving null are (roughly speaking) false by
definition.
We shall study precise meaning of comparisons with nulls later
©Silberschatz, Korth and Sudarshan2.56Database System Concepts - 5th Edition, Oct 5, 2006
Outer Join
Outer Join
Example
Example
Relation loan
Relation borrower
customer_name loan_number
Jones
Smith
Hayes
L-170
L-230
L-155
3000
4000
1700
loan_number amount
L-170
L-230
L-260
branch_name
Downtown
Redwood
Perryridge
©Silberschatz, Korth and Sudarshan2.57Database System Concepts - 5th Edition, Oct 5, 2006
Outer Join
Outer Join
Example
Example
Join
loan borrower
loan_number amount
L-170
L-230
3000
4000
customer_name
Jones
Smith
branch_name
Downtown
Redwood
Jones
Smith
null
loan_number amount
L-170
L-230
L-260
3000
4000
1700
customer_namebranch_name
Downtown
Redwood
Perryridge
Left Outer Join
loan borrower
©Silberschatz, Korth and Sudarshan2.58Database System Concepts - 5th Edition, Oct 5, 2006
Outer Join
Outer Join
Example
Example
loan_number amount
L-170
L-230
L-155
3000
4000
null
customer_name
Jones
Smith
Hayes
branch_name
Downtown
Redwood
null
loan_number amount
L-170
L-230
L-260
L-155
3000
4000
1700
null
customer_name
Jones
Smith
null
Hayes
branch_name
Downtown
Redwood
Perryridge
null
Full Outer Join
loan borrower
Right Outer Join
loan borrower
©Silberschatz, Korth and Sudarshan2.59Database System Concepts - 5th Edition, Oct 5, 2006
Null Values
Null Values
It is possible for tuples to have a null value, denoted by null, for some
of their attributes
null signifies an unknown value or that a value does not exist.
The result of any arithmetic expression involving null is null.
Aggregate functions simply ignore null values (as in SQL)
For duplicate elimination and grouping, null is treated like any other
value, and two nulls are assumed to be the same (as in SQL)
©Silberschatz, Korth and Sudarshan2.60Database System Concepts - 5th Edition, Oct 5, 2006
Null Values
Null Values
Comparisons with null values return the special truth value:
unknown
If false was used instead of unknown, then not (A < 5)
would not be equivalent to A >= 5
Three-valued logic using the truth value unknown:
OR: (unknown or true) = true,
(unknown or false) = unknown
(unknown or unknown)= unknown
AND: (true and unknown)= unknown,
(false and unknown)= false,
(unknown and unknown)= unknown
NOT: (not unknown)= unknown
In SQL “Pis unknown evaluates to true if predicate Pevaluates
to unknown
Result of select predicate is treated as false if it evaluates to
unknown
©Silberschatz, Korth and Sudarshan2.61Database System Concepts - 5th Edition, Oct 5, 2006
Modification of the Database
Modification of the Database
The content of the database may be modified using the following
operations:
Deletion
Insertion
Updating
All these operations are expressed using the assignment
operator.
©Silberschatz, Korth and Sudarshan2.62Database System Concepts - 5th Edition, Oct 5, 2006
Deletion
Deletion
A delete request is expressed similarly to a query, except
instead of displaying tuples to the user, the selected tuples are
removed from the database.
Can delete only whole tuples; cannot delete values on only
particular attributes
A deletion is expressed in relational algebra by:
rrE
where ris a relation and Eis a relational algebra query.
©Silberschatz, Korth and Sudarshan2.63Database System Concepts - 5th Edition, Oct 5, 2006
Deletion Examples
Deletion Examples
Delete all account records in the Perryridge branch.
Delete all accounts at branches located in Needham.
r
1
σ
branch_city = “Needham”
(account branch )
r
2
account_number,branch_name, balance
(r
1
)
r
3
customer_name, account_number
(r
2
depositor)
account account – r
2
depositor depositor – r
3
Delete all loan records with amount in the range of 0 to 50
loan loan
σ
amount
0
and amount
50
(loan)
account account
σ
branch_name = “Perryridge
(account )
©Silberschatz, Korth and Sudarshan2.64Database System Concepts - 5th Edition, Oct 5, 2006
Insertion
Insertion
To insert data into a relation, we either:
specify a tuple to be inserted
write a query whose result is a set of tuples to be inserted
in relational algebra, an insertion is expressed by:
r rE
where ris a relation and Eis a relational algebra expression.
The insertion of a single tuple is expressed by letting Ebe a constant
relation containing one tuple.
©Silberschatz, Korth and Sudarshan2.65Database System Concepts - 5th Edition, Oct 5, 2006
Insertion Examples
Insertion Examples
Insert information in the database specifying that Smith has $1200 in
account A-973 at the Perryridge branch.
Provide as a gift for all loan customers in the Perryridge
branch, a $200 savings account. Let the loan number serve
as the account number for the new savings account.
account account {(“A-973”, “Perryridge”, 1200)}
depositor depositor {(“Smith”, “A-973”)}
r
1
(σ
branch_name = “Perryridge”
(borrower loan))
account account
loan_number,
branch_name,
200
(r
1
)
depositor depositor
customer_name, loan_number
(r
1
)
©Silberschatz, Korth and Sudarshan2.66Database System Concepts - 5th Edition, Oct 5, 2006
Updating
Updating
A mechanism to change a value in a tuple without charging all values in
the tuple
Use the generalized projection operator to do this task
Each F
i
is either
the I
th
attribute of r, if the I
th
attribute is not updated, or,
if the attribute is to be updated F
i
is an expression, involving only
constants and the attributes of r, which gives the new value for the
attribute
)(
,,,,
21
rr
l
FFF K
©Silberschatz, Korth and Sudarshan2.67Database System Concepts - 5th Edition, Oct 5, 2006
Update Examples
Update Examples
Make interest payments by increasing all balances by 5 percent.
Pay all accounts with balances over $10,000 6 percent interest
and pay all others 5 percent
account
account_number, branch_name, balance * 1.06
(σ
BAL
>
10000
(account ))
account_number, branch_name, balance * 1.05
(σ
BAL
10000
(account))
account
account_number, branch_name, balance * 1.05
(account)
Database System Concepts, 5
th
Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
End of Chapter 2
End of Chapter 2
©Silberschatz, Korth and Sudarshan2.69Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.3. The
Figure 2.3. The branch
branch relation
relation
©Silberschatz, Korth and Sudarshan2.70Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.6: The
Figure 2.6: The loan
loan relation
relation
©Silberschatz, Korth and Sudarshan2.71Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.7: The
Figure 2.7: The borrower
borrower relation
relation
©Silberschatz, Korth and Sudarshan2.72Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.9
Figure 2.9
Result of
Result of σ
branch_name = “Perryridge”
(loan)
©Silberschatz, Korth and Sudarshan2.73Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.10:
Figure 2.10:
Loan number and the amount of the loan
Loan number and the amount of the loan
©Silberschatz, Korth and Sudarshan2.74Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.11: Names of all customers who
Figure 2.11: Names of all customers who
have either an account or an loan
have either an account or an loan
©Silberschatz, Korth and Sudarshan2.75Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.12:
Figure 2.12:
Customers with an account but no loan
Customers with an account but no loan
©Silberschatz, Korth and Sudarshan2.76Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.13: Result of
Figure 2.13: Result of borrower
borrower |X|
|X| loan
loan
©Silberschatz, Korth and Sudarshan2.77Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.14
Figure 2.14
©Silberschatz, Korth and Sudarshan2.78Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.15
Figure 2.15
©Silberschatz, Korth and Sudarshan2.79Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.16
Figure 2.16
©Silberschatz, Korth and Sudarshan2.80Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.17
Figure 2.17
Largest account balance in the bank
Largest account balance in the bank
©Silberschatz, Korth and Sudarshan2.81Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.18: Customers who live on the
Figure 2.18: Customers who live on the
same street and in the same city as
same street and in the same city as
Smith
Smith
©Silberschatz, Korth and Sudarshan2.82Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.19: Customers with both an
Figure 2.19: Customers with both an
account and a loan at the bank
account and a loan at the bank
©Silberschatz, Korth and Sudarshan2.83Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.20
Figure 2.20
©Silberschatz, Korth and Sudarshan2.84Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.21
Figure 2.21
©Silberschatz, Korth and Sudarshan2.85Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.22
Figure 2.22
©Silberschatz, Korth and Sudarshan2.86Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.23
Figure 2.23
©Silberschatz, Korth and Sudarshan2.87Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.24: The
Figure 2.24: The credit_info
credit_info relation
relation
©Silberschatz, Korth and Sudarshan2.88Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.25
Figure 2.25
©Silberschatz, Korth and Sudarshan2.89Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.26: The
Figure 2.26: The pt_works
pt_works relation
relation
©Silberschatz, Korth and Sudarshan2.90Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.27
Figure 2.27
The
The pt_works
pt_works relation after regrouping
relation after regrouping
©Silberschatz, Korth and Sudarshan2.91Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.28
Figure 2.28
©Silberschatz, Korth and Sudarshan2.92Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.29
Figure 2.29
©Silberschatz, Korth and Sudarshan2.93Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.30
Figure 2.30
The
The employee
employee and
and ft_works relations
ft_works relations
©Silberschatz, Korth and Sudarshan2.94Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.31
Figure 2.31
©Silberschatz, Korth and Sudarshan2.95Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.32
Figure 2.32
©Silberschatz, Korth and Sudarshan2.96Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.33
Figure 2.33
©Silberschatz, Korth and Sudarshan2.97Database System Concepts - 5th Edition, Oct 5, 2006
Figure 2.34
Figure 2.34